Question

Solve the equation for x, accurate to three decimal places: ex − 2e−x = 1.

x = 0.301
x = 0.693
x = 0.151
x = 0.347

Answer 1

my guess
\[e^x-2e^{-x}=1\\
x=\ln(2)\]

Answer 2

did you try to factor it ?

Answer 3

i do not know how to do this at all

Answer 4

\[e ^{x}-2e^{-x}=1\]\[e ^{x}-2e^{-x}-1=0\]

Answer 5

o ok

Answer 6

x = 0.347 is this it

Answer 7

nope, but as sattellite suggested, if you plug in x = ln 2 , that works

Answer 8

\[e^x -2e^{-x} = 1\]\[\frac{e^x}{e^x} (e^x -2e^{-x})=1\]\[\frac{1}{e^x} (e^{2x} -2)=1\]This way might work as well, no?

Answer 9

basically I factored out an \(e^{-x}\) which can be rewritten as \(\dfrac{e^x}{e^x}\)

Answer 10

o ok so it is B

Answer 11

Then I multiplied the "top" by \(e^x\)

Answer 12

yes

Answer 13

@Jhannybean did you get x = ln 2 as your solution ?

Answer 14

I haven't solved it yet :P

Answer 15

oh :)

Answer 16

\[\frac{1}{e^x} (e^{2x} -2)=1\]\[\frac{e^{2x}}{e^x} -\frac{2}{e^x} = 1\]\[e^{2x}-2=e^x\]I'm not quite sure how to solve this

Answer 17

ohh i see now

Answer 18

i misspoke earlier, you might be able to factor this

Answer 19

Oh... \[e^{2x} -e^x =2\]\[e^x = 2\]\[\log_e (e^x) = \log_e (2)\]\[x= \log_e (2) = \ln (2)\]

Answer 20

@satellite73 is this how you would have solved this?

Answer 21

A is it right

Answer 22

input \(\ln (2)\) into your calculator, what do you get?

Answer 23

Yeah that's the same thing I did :P

Answer 24

ok good we got the same answer :)

Answer 25

start with \[e^{2x}-2=e^x\] and then
\[e^{2x}-e^x-2=0\] treat it as a quadratic

Answer 26

so A is right

Answer 27

no it is c

Answer 28

e^x - 2e^(-x) = 1

e^x - 2/e^x = 1

multiply both sides by e^x, which will cancel the e^x in the denominator

e^x*e^x - 2 = e^x

e^(x+x) - e^x -2 = 0

e^(2x) - e^x - 2 = 0

(e^x)^2 - 1*(e^x) -2 = 0

(e^x - 2 )*( e^x + 1 ) = 0

e^x - 2 = 0 , e^x + 1 = 0

e^x = 2 , e^x = -1 (no solution)

ln(e^x) = ln 2

x = ln 2